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Evaluate the following double integral

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Change the order of integration and evaluate the following double integral:

    [tex]I = {\int_0^{1} \left({\int\limits_{y}^{1}
    30 y\sqrt{1+x^3} \mathrm{d}x }\right) {\mathrm{d}y} [/tex]


    So thenn i did

    [tex] = 30 \int_0^{1} \sqrt{1+x^3} \left({\int_0^{x} y \mathrm{d}y}\right) \mathrm{d}x [/tex]

    [tex]= 30 \int_0^{1} \sqrt{1+x^3} \left(\frac{x^2}{2} \right) \mathrm{d}x \end{align}[/tex]

    using integration by parts....

    for [tex] \sqrt{1+x^3} [/tex]
    [tex] let u = \sqrt{1+x^3} \qquad du= \frac{1}{2} \left(\sqrt{1+x^3}\right) 3x^2 = \frac{3x^2}{2\sqrt{1+x^3}} \qquad dv = dx \qquad v = x [/tex]

    Thus!

    [tex] = x \sqrt{1+x^3} - \int \frac{3x^3}{2\sqrt{1+x^3}} \mathrm{d}x [/tex]


    after that.... i have no clue what to do. a lil help? thanks! :smile:
    am i on the right track though?
     
  2. jcsd
  3. May 23, 2010 #2

    gabbagabbahey

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    Re: Integration!

    Why use IBP at all? What is [itex]\frac{d}{dx}(1+x^3)^{3/2}[/itex]?:wink:
     
  4. May 24, 2010 #3
    Re: Integration!

    don't we need to worry bout what's inside the bracket? when differentiating? :uhh:
     
  5. May 24, 2010 #4

    gabbagabbahey

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    Re: Integration!

    Of course, use the chain rule.
     
  6. May 24, 2010 #5
    Re: Integration!

    OHHHHHHHHH!!!!!!!!!!!!!!!!!
    ahh dear.. i sure do love to make things complicated.. :rofl:

    Thanks heaps! and there i was looking at tht question for hrs....:blushing:
     
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